(a + b + c)³ = [(a+b) + c]³ = (a+b)³ +c³ +3(a+b)c(a+b+c) =
= a³ + b³ + c³ + 3ab(a + b) + 3(a+b)c(a+b+c) =
= a³ + b³ + c³ + 3(a+b)(ab + ac+bc+c²) =
= a³ + b³ + c³ + 3(a+b)[(ab+ac) +(bc+c²)] =
= a³ + b³ + c³ + 3(a+b)[a(b + c) + c(b + c)] =
= a³ + b³ + c³ + 3(a+b)(b+c)(a+c).
Prin urmare :
(a + b + c)³ = a³ + b³ + c³ + 3(a+b)(b+c)(a+c) ⇒
⇒ (a + b + c)³ - 3(a+b)(b+c)(a+c) = a³ + b³ + c³ ⇒
⇒a³ + b³ + c³= (a + b + c)³ - 3(a+b)(b+c)(a+c).